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Full Version: An alternate power series representation for ln(x)
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This proof involves the use of a new operator:

and it's inverse:

and the little differential operator:

see here for more: http://math.eretrandre.org/tetrationforu...40#pid5740

I'll use these specifically:

and

The proof starts out by first proving:

first give the power series representation of e^x

And given:

We take the ln of e^x to get an infinite series of deltations, if:
represents a series of deltations

then

and therefore if we let x = e^x

and now we have an infinite lowered polynomial which when little differentiated equals itself. It's the little polynomial equivalent of e^x's perfect taylor series.

therefore:

and using the chain rule:
where f'(x) is taken to mean the little derivative of f(x).

we get the result:

and now we can solve for the k'th little derivative of ln(x) using the little power rule, k E N

and so if the little derivative Taylor series is given by:

where is the n'th little derivative of .

we can take the little derivative taylor series of ln(x) centered about 1.
The first term is equal to 0, so I'll start the series from n = 1.

Here it is in two steps:

now let's plug this in our formula for ; it works for an infinite sum because is commutative and associative.

now since:

now take the lns away and

and now if we let x = ln(x) we get:

And there, that's it. I haven't been able to check if this series converges, I don't have many convergence tests. I'm just sure that the algebra is right. I haven't tried computing it. In my gut it doesn't look like it will converge, but I have faith.
Doing some tests it converges for 1, 2, 4, 5 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works.

Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of ) I find the normal taylor series of ln(x):

so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0.
(05/07/2011, 08:41 PM)JmsNxn Wrote: [ -> ]This proof involves the use of a new operator:

and it's inverse:

and the little differential operator:

(The notation is more unambiguous than in your previous thread )

But your operator can be expressed with the classical differentiation, see:

Or purely functional with the composition operation :

PS: when you write ln with backslash in front:
Code:
$$\ln(x)$$
you get a better ln-typesetting.
Yeah, I was aware of a direct relation to differentiation:

And I've extended the definition of the series to:

And I've found a beautiful return:

Converges for values as high as 30. My computer overflows before it stops converging.

I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x)

edit:
sadly, doesn't converge for values less than 1
(05/07/2011, 11:20 PM)JmsNxn Wrote: [ -> ]

But this follows directly from the definition of the logarithm, by the following equivalent transformations:

Quote:sadly, doesn't converge for values less than 1
The series converges for , hence your series converges for , this should include all values ?
I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the went to in the denominator. Otherwise, though, that's a nice proof.
(05/08/2011, 07:54 PM)JmsNxn Wrote: [ -> ]I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the went to in the denominator. Otherwise, though, that's a nice proof.

Oh thats just:
(05/08/2011, 08:28 PM)bo198214 Wrote: [ -> ]Oh thats just:

Oh that's so simple! Thanks for the help.